Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:640000000")
using namespace std;
const double EPS = 1e-9;
const int INF = 0x7f7f7f7f;
const double PI = acos(-1.0);
template <class T>
inline T gcd(T a, T b) {
  return (b) == 0 ? (a) : gcd((b), ((a) % (b)));
}
template <class T>
inline T lcm(T a, T b) {
  return ((a) / gcd((a), (b)) * (b));
}
template <typename T>
string NumberToString(T Number) {
  ostringstream second;
  second << Number;
  return second.str();
}
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
int t, n, m;
int arr[105][105];
int cur[105][105];
int row[105];
int col[105];
bool ok() {
  for (int i = 1; i <= (n); i++)
    for (int j = 1; j <= (m); j++)
      if (arr[i][j]) return false;
  return true;
}
int findminrow(int pos) {
  int mini = INF;
  for (int i = 1; i <= (m); i++) mini = min(mini, arr[pos][i]);
  for (int i = 1; i <= (m); i++) arr[pos][i] -= mini;
  return mini;
}
int findmincol(int pos) {
  int mini = INF;
  for (int i = 1; i <= (n); i++) mini = min(mini, arr[i][pos]);
  for (int i = 1; i <= (n); i++) arr[i][pos] -= mini;
  return mini;
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    for (int i = 1; i <= (n); i++)
      for (int j = 1; j <= (m); j++) scanf("%d", &arr[i][j]);
    memset(cur, 0, sizeof(cur));
    int ans = 0;
    if (n <= m) {
      for (int i = 1; i <= (n); i++) row[i] = findminrow(i), ans += row[i];
      for (int i = 1; i <= (m); i++) col[i] = findmincol(i), ans += col[i];
    } else {
      for (int i = 1; i <= (m); i++) col[i] = findmincol(i), ans += col[i];
      for (int i = 1; i <= (n); i++) row[i] = findminrow(i), ans += row[i];
    }
    if (ok()) {
      printf("%d\n", ans);
      for (int i = 1; i <= (n); i++) {
        for (int j = 1; j <= (row[i]); j++) printf("row %d\n", i);
      }
      for (int i = 1; i <= (m); i++) {
        for (int j = 1; j <= (col[i]); j++) printf("col %d\n", i);
      }
    } else
      puts("-1");
  }
  return 0;
}